# A static equilibrium problem.

1. Sep 22, 2008

### akan

1. The problem statement, all variables and given/known data
A uniform board of length L and weight W is suspended between two vertical walls by ropes of length L/2 each. When a weight w is placed on the left end of the board, it assumes the configuration shown in the figure.

http://img260.imageshack.us/img260/3992/rw1234ok6.jpg [Broken]

Find the weight w in terms of the board weight W.

2. Relevant equations
If the left rope corresponds to T_1, and the right rope to T_2, left and down is negative, right and up is positive, then:

Sum(F_x) = -T_1 sin(35) + T_2 sin(60) = 0
Sum(F_y) = T_1 cos(35) + T_2 cos (60) - W - w = 0
Sum(T_z) = - W(L/2) cos(9.2) + T_2 L sin(20.8 [just calculated this from graphical drawing]) = 0 [pivot at w].

3. The attempt at a solution
eq1. T_2 sin(60) = T_1 sin(35)
eq1. T_2 = T_1 sin(35)/sin(60)

eq2. T_1 (cos(35) + sin(35)cos(60)/sin(60)) - W = w

eq3. T_2 L sin(20.8) = W (L/2) cos(9.2)
eq3. T_2 sin(20.8) = W (1/2) cos(9.2)
eq3. T_1 sin(35)sin(20.8)/sin(60) = W (1/2) cos(9.2)
eq3. T_1 = [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)]

eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W * .4274425413] / [.203680986] * [1.150306554] = w
eq2 w = 2.41 W

The correct answer is 1.42 W. What did I do wrong?

Last edited by a moderator: May 3, 2017
2. Sep 22, 2008

### tiny-tim

Hi akan!

You left out the - W.

(though it might hav been quicker

i] to find T1 in terms of T2 instead of vice versa

ii] to take moments about the point where the lines of the two ropes meet )

3. Sep 22, 2008

### akan

Thank you!

1) Yes, you're right, that's easier.
2) Uh... If I did that, how would I account for the torque due to W and w? Where would the moment arm be? Wouldn't that make more unknown variables, too?

4. Sep 23, 2008

### tiny-tim

Well, you know that the combined torque of W and w must go through that point … and they're both vertical, so you just balance the left and right amounts, like an uneven see-saw.

(you get the same equation in the end, of course …*but it's all in one geometrical diagram, so it's easier to check what you're doing)

5. Sep 23, 2008

### akan

Can you please explain how to express torque if I set the pivot point in such a way that it does not even lie along the level arm? This is something that neither my textbook nor my professor seem to be able to explain clearly. Thanks.

6. Sep 24, 2008

### tiny-tim

(btw, the lever arm (stupid name, IMO) isn't the line of the force … it's the distance from the pivot point to the line of the force).

You can set the pivot point anywhere

(The pivot point I suggested has the advantage that it makes two of the lever arms (and therefore the torques) zero! )

torque = force times perpendicular distance …

that is, the perpendicular distance (shortest distance) from the pivot point to the line of application of the force.

This extract from the PF Library may help:
Any questions?

7. Sep 25, 2008

### akan

Last edited by a moderator: May 3, 2017
8. Sep 26, 2008

### tiny-tim

Hi akan!
Yes!
You only need the horizontal distances from P (the pivot) to W and w … the forces w and W are vertical, so the "lever arm" (horrible phrase) is horizontal.

The horizontal distance to w is z1sin35º, and the horizontal distance to W is (L/2)cos9.2º minus that.

9. Sep 26, 2008

### akan

Shouldn't the distances be perpendicular to the lever arm (please also suggest a better term), because torque is r x F = rF sin theta, where theta is the angle between the two vectors? And please explain how'd you get the horizontal distance to W from the pivot. Thanks.

10. Sep 26, 2008

### tiny-tim

The distance is the lever arm.

In r x F = rF sin theta, F is the force (along the "line of force"), and r is the lever arm.

I just call r the "perpendicular distance".

The horizontal distance to w is z1sin35º (and you can get z1 from the sine rule), and the horizontal distance to W is (L/2)cos9.2º minus that.

11. Sep 26, 2008

### akan

Can you please draw a picture from which it would be clear where the (L/2) cos 9.2 comes from? Thank you.

12. Sep 26, 2008

### tiny-tim

erm … it's your picture …

(L/2) cos 9.2º is the horizontal distance from W (the centre of the board) to w.

13. Sep 26, 2008

### akan

Ah. I see what you're saying. But I need the horizontal distance from W to the pivot, not from W to w.

14. Sep 26, 2008

### tiny-tim

grrr! :grumpy: